Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 4;
const int LN = 30;
int n;
int a[N];
int b[N];
bool done[N];
vector<pair<int, int> > va;
vector<pair<int, int> > vb;
int solve(int arr[], vector<pair<int, int> > &res) {
  for (int i = 1; i <= n; ++i) {
    done[i] = 0;
  }
  for (int i = LN - 1; i >= 0; --i) {
    int idx = -1;
    for (int j = 1; j <= n; ++j) {
      if (!done[j] && arr[j] & (1 << i)) {
        idx = j;
        break;
      }
    }
    if (idx != -1) {
      done[idx] = 1;
      for (int j = 1; j <= n; ++j) {
        if (!done[j] && (arr[j] & (1 << i))) {
          arr[j] ^= arr[idx];
          res.emplace_back(make_pair(j, idx));
        }
      }
    }
  }
  int idx = 1;
  for (int i = LN - 1; i >= 0; --i) {
    for (int j = idx; j <= n; ++j) {
      if (arr[j] & (1 << i)) {
        if (j != idx) {
          arr[j] ^= arr[idx];
          arr[idx] ^= arr[j];
          arr[j] ^= arr[idx];
          res.emplace_back(make_pair(j, idx));
          res.emplace_back(make_pair(idx, j));
          res.emplace_back(make_pair(j, idx));
        }
        ++idx;
        break;
      }
    }
  }
  for (int i = idx; i <= n; ++i) {
    arr[i] ^= arr[i];
    res.emplace_back(make_pair(i, i));
  }
  return idx - 1;
}
set<int> get(int val) {
  int lol = 1 << LN;
  int idx = 1;
  set<int> res;
  res.clear();
  for (int i = LN - 1; i >= 0; --i) {
    if (val & (1 << i)) {
      while (a[idx] & lol) {
        ++idx;
      }
      if (a[idx] & (1 << i)) {
        val ^= a[idx];
        res.insert(idx);
      } else {
        printf("-1\n");
        exit(0);
      }
    }
    lol |= 1 << i;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", b + i);
  }
  int b1 = solve(a, va);
  int b2 = solve(b, vb);
  for (int i = 1; i <= b2; ++i) {
    auto it = get(b[i]);
    if (it.find(i) == it.end()) {
      it.insert(i);
    } else {
      it.erase(i);
    }
    for (auto x : it) {
      a[i] ^= a[x];
      va.emplace_back(make_pair(i, x));
    }
  }
  for (int i = b2 + 1; i <= n; ++i) {
    a[i] ^= a[i];
    va.emplace_back(make_pair(i, i));
  }
  printf("%d\n", int(va.size()) + int(vb.size()));
  for (auto it : va) {
    printf("%d %d\n", it.first, it.second);
  }
  reverse(vb.begin(), vb.end());
  for (auto it : vb) {
    printf("%d %d\n", it.first, it.second);
  }
}